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输入有序数组(模板题)<span><svg xmlns="http://www.w3.org/2000/svg" aria-hidden="true" focusable="false" x="0px" y="0px" viewBox="0 0 100 100" width="15" height="15" class="icon outbound"><path fill="currentColor" d="M18.8,85.1h56l0,0c2.2,0,4-1.8,4-4v-32h-8v28h-48v-48h28v-8h-32l0,0c-2.2,0-4,1.8-4,4v56C14.8,83.3,16.6,85.1,18.8,85.1z"></path> <polygon fill="currentColor" points="45.7,48.7 51.3,54.3 77.2,28.5 77.2,37.2 85.2,37.2 85.2,14.9 62.8,14.9 62.8,22.9 71.5,22.9"></polygon></svg> <span class="sr-only">(opens new window)</span></span></a></h4> <p>给你一个下标从 <strong>1</strong> 开始的整数数组 <code>numbers</code> ，该数组已按 <strong>非递减顺序排列</strong> ，请你从数组中找出满足相加之和等于目标数 <code>target</code> 的两个数。如果设这两个数分别是 <code>numbers[index1]</code> 和 <code>numbers[index2]</code> ，则 <code>1 &lt;= index1 &lt; index2 &lt;= numbers.length</code> 。</p> <p>以长度为 2 的整数数组 <code>[index1, index2]</code> 的形式返回这两个整数的下标 <code>index1</code> 和 <code>index2</code>。</p> <p>你可以假设每个输入 <strong>只对应唯一的答案</strong> ，而且你 <strong>不可以</strong> 重复使用相同的元素。</p> <p>你所设计的解决方案必须只使用常量级的额外空间。</p> <p><strong>示例 1：</strong></p> <div class="language- line-numbers-mode"><pre class="language-text"><code>输入：numbers = [2,7,11,15], target = 9
输出：[1,2]
解释：2 与 7 之和等于目标数 9 。因此 index1 = 1, index2 = 2 。返回 [1, 2] 。
</code></pre> <div class="line-numbers-wrapper"><span class="line-number">1</span><br><span class="line-number">2</span><br><span class="line-number">3</span><br></div></div><p><strong>示例 2：</strong></p> <div class="language- line-numbers-mode"><pre class="language-text"><code>输入：numbers = [2,3,4], target = 6
输出：[1,3]
解释：2 与 4 之和等于目标数 6 。因此 index1 = 1, index2 = 3 。返回 [1, 3] 。
</code></pre> <div class="line-numbers-wrapper"><span class="line-number">1</span><br><span class="line-number">2</span><br><span class="line-number">3</span><br></div></div><p><strong>示例 3：</strong></p> <div class="language- line-numbers-mode"><pre class="language-text"><code>输入：numbers = [-1,0], target = -1
输出：[1,2]
解释：-1 与 0 之和等于目标数 -1 。因此 index1 = 1, index2 = 2 。返回 [1, 2] 。
</code></pre> <div class="line-numbers-wrapper"><span class="line-number">1</span><br><span class="line-number">2</span><br><span class="line-number">3</span><br></div></div><p><strong>提示：</strong></p> <ul><li><code>2 &lt;= numbers.length &lt;= 3 * 104</code></li> <li><code>-1000 &lt;= numbers[i] &lt;= 1000</code></li> <li><code>numbers</code> 按 <strong>非递减顺序</strong> 排列</li> <li><code>-1000 &lt;= target &lt;= 1000</code></li> <li><strong>仅存在一个有效答案</strong></li></ul> <h5 id="解题思路"><a href="#解题思路" class="header-anchor">#</a> 解题思路</h5> <div class="language-shell line-numbers-mode"><pre class="language-shell"><code><span class="token number">1</span>. 双指针夹逼，条件：数组升序，左右和
<span class="token number">2</span>. 字典，来一个存一个，如果找到target-i,获取返回
</code></pre> <div class="line-numbers-wrapper"><span class="line-number">1</span><br><span class="line-number">2</span><br></div></div><h5 id="完整代码"><a href="#完整代码" class="header-anchor">#</a> 完整代码</h5> <ul><li>双指针</li></ul> <div class="language-js line-numbers-mode"><pre class="language-js"><code><span class="token comment">/**
 * @param {number[]} numbers
 * @param {number} target
 * @return {number[]}
 */</span>
<span class="token keyword">var</span> <span class="token function-variable function">twoSum</span> <span class="token operator">=</span> <span class="token keyword">function</span><span class="token punctuation">(</span><span class="token parameter">numbers<span class="token punctuation">,</span> target</span><span class="token punctuation">)</span> <span class="token punctuation">{</span>

    <span class="token comment">// 使用双指针扫描</span>

    <span class="token comment">// 1. 滑动窗口 i,j方向相同</span>

    <span class="token comment">// 2 .i，j相向</span>

    <span class="token comment">// 这里 i，j相向</span>

    <span class="token comment">// 因为数组升序，i一开始最小，j一开始最大</span>
    <span class="token comment">// i+j&gt;target i不动，j左移 ==&gt; target 变小  ，如果满足，直接返回，如果&lt;target 说明i要右移 target变大</span>

    <span class="token comment">// 夹逼最后找到解</span>

    <span class="token keyword">let</span> j<span class="token operator">=</span>numbers<span class="token punctuation">.</span>length<span class="token operator">-</span><span class="token number">1</span>
    <span class="token keyword">for</span><span class="token punctuation">(</span><span class="token keyword">let</span> i<span class="token operator">=</span><span class="token number">0</span><span class="token punctuation">;</span>i<span class="token operator">&lt;</span>numbers<span class="token punctuation">.</span>length<span class="token punctuation">;</span>i<span class="token operator">++</span><span class="token punctuation">)</span><span class="token punctuation">{</span>
        <span class="token keyword">while</span><span class="token punctuation">(</span>i<span class="token operator">&lt;</span>j<span class="token operator">&amp;&amp;</span>numbers<span class="token punctuation">[</span>i<span class="token punctuation">]</span><span class="token operator">+</span>numbers<span class="token punctuation">[</span>j<span class="token punctuation">]</span><span class="token operator">&gt;</span>target<span class="token punctuation">)</span><span class="token punctuation">{</span>
            j<span class="token operator">--</span>
        <span class="token punctuation">}</span>

        <span class="token keyword">if</span><span class="token punctuation">(</span>i<span class="token operator">&lt;</span>j<span class="token operator">&amp;&amp;</span>numbers<span class="token punctuation">[</span>i<span class="token punctuation">]</span><span class="token operator">+</span>numbers<span class="token punctuation">[</span>j<span class="token punctuation">]</span><span class="token operator">===</span>target<span class="token punctuation">)</span><span class="token punctuation">{</span>
            <span class="token keyword">return</span> <span class="token punctuation">[</span>i<span class="token operator">+</span><span class="token number">1</span><span class="token punctuation">,</span>j<span class="token operator">+</span><span class="token number">1</span><span class="token punctuation">]</span>
        <span class="token punctuation">}</span>
    <span class="token punctuation">}</span>

<span class="token punctuation">}</span><span class="token punctuation">;</span>
</code></pre> <div class="line-numbers-wrapper"><span class="line-number">1</span><br><span class="line-number">2</span><br><span class="line-number">3</span><br><span class="line-number">4</span><br><span class="line-number">5</span><br><span class="line-number">6</span><br><span class="line-number">7</span><br><span class="line-number">8</span><br><span class="line-number">9</span><br><span class="line-number">10</span><br><span class="line-number">11</span><br><span class="line-number">12</span><br><span class="line-number">13</span><br><span class="line-number">14</span><br><span class="line-number">15</span><br><span class="line-number">16</span><br><span class="line-number">17</span><br><span class="line-number">18</span><br><span class="line-number">19</span><br><span class="line-number">20</span><br><span class="line-number">21</span><br><span class="line-number">22</span><br><span class="line-number">23</span><br><span class="line-number">24</span><br><span class="line-number">25</span><br><span class="line-number">26</span><br><span class="line-number">27</span><br><span class="line-number">28</span><br><span class="line-number">29</span><br><span class="line-number">30</span><br><span class="line-number">31</span><br><span class="line-number">32</span><br></div></div><ul><li>字典</li></ul> <div class="language-js line-numbers-mode"><pre class="language-js"><code>    <span class="token comment">// 可以使用字典</span>
    <span class="token comment">// 每来一个元素，计算它的另一半（target-i）</span>
    <span class="token comment">// 看是否已经存在字典中</span>
    <span class="token keyword">let</span> m<span class="token operator">=</span><span class="token keyword">new</span> <span class="token class-name">Map</span><span class="token punctuation">(</span><span class="token punctuation">)</span>
    <span class="token keyword">for</span><span class="token punctuation">(</span><span class="token keyword">let</span> i<span class="token operator">=</span><span class="token number">0</span><span class="token punctuation">;</span>i<span class="token operator">&lt;</span>numbers<span class="token punctuation">.</span>length<span class="token punctuation">;</span>i<span class="token operator">++</span><span class="token punctuation">)</span><span class="token punctuation">{</span>

        <span class="token keyword">let</span> j<span class="token operator">=</span>target<span class="token operator">-</span>numbers<span class="token punctuation">[</span>i<span class="token punctuation">]</span>
        <span class="token keyword">if</span><span class="token punctuation">(</span>m<span class="token punctuation">.</span><span class="token function">has</span><span class="token punctuation">(</span>j<span class="token punctuation">)</span><span class="token punctuation">)</span><span class="token punctuation">{</span>
            <span class="token keyword">return</span> <span class="token punctuation">[</span>m<span class="token punctuation">.</span><span class="token function">get</span><span class="token punctuation">(</span>j<span class="token punctuation">)</span><span class="token punctuation">,</span>i<span class="token operator">+</span><span class="token number">1</span><span class="token punctuation">]</span>
        <span class="token punctuation">}</span><span class="token keyword">else</span><span class="token punctuation">{</span>
            m<span class="token punctuation">.</span><span class="token function">set</span><span class="token punctuation">(</span>numbers<span class="token punctuation">[</span>i<span class="token punctuation">]</span><span class="token punctuation">,</span>i<span class="token operator">+</span><span class="token number">1</span><span class="token punctuation">)</span>
        <span class="token punctuation">}</span>
        
    <span class="token punctuation">}</span>
</code></pre> <div class="line-numbers-wrapper"><span class="line-number">1</span><br><span class="line-number">2</span><br><span class="line-number">3</span><br><span class="line-number">4</span><br><span class="line-number">5</span><br><span class="line-number">6</span><br><span class="line-number">7</span><br><span class="line-number">8</span><br><span class="line-number">9</span><br><span class="line-number">10</span><br><span class="line-number">11</span><br><span class="line-number">12</span><br><span class="line-number">13</span><br><span class="line-number">14</span><br></div></div><h4 id="_15-三数之和"><a href="#_15-三数之和" class="header-anchor">#</a> <a href="https://leetcode.cn/problems/3sum/description/" target="_blank" rel="noopener noreferrer">15. 三数之和<span><svg xmlns="http://www.w3.org/2000/svg" aria-hidden="true" focusable="false" x="0px" y="0px" viewBox="0 0 100 100" width="15" height="15" class="icon outbound"><path fill="currentColor" d="M18.8,85.1h56l0,0c2.2,0,4-1.8,4-4v-32h-8v28h-48v-48h28v-8h-32l0,0c-2.2,0-4,1.8-4,4v56C14.8,83.3,16.6,85.1,18.8,85.1z"></path> <polygon fill="currentColor" points="45.7,48.7 51.3,54.3 77.2,28.5 77.2,37.2 85.2,37.2 85.2,14.9 62.8,14.9 62.8,22.9 71.5,22.9"></polygon></svg> <span class="sr-only">(opens new window)</span></span></a></h4> <p>给你一个整数数组 <code>nums</code> ，判断是否存在三元组 <code>[nums[i], nums[j], nums[k]]</code> 满足 <code>i != j</code>、<code>i != k</code> 且 <code>j != k</code> ，同时还满足 <code>nums[i] + nums[j] + nums[k] == 0</code> 。请</p> <p>你返回所有和为 <code>0</code> 且不重复的三元组。</p> <p>**注意：**答案中不可以包含重复的三元组。</p> <p><strong>示例 1：</strong></p> <div class="language- line-numbers-mode"><pre class="language-text"><code>输入：nums = [-1,0,1,2,-1,-4]
输出：[[-1,-1,2],[-1,0,1]]
解释：
nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0 。
nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0 。
nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0 。
不同的三元组是 [-1,0,1] 和 [-1,-1,2] 。
注意，输出的顺序和三元组的顺序并不重要。
</code></pre> <div class="line-numbers-wrapper"><span class="line-number">1</span><br><span class="line-number">2</span><br><span class="line-number">3</span><br><span class="line-number">4</span><br><span class="line-number">5</span><br><span class="line-number">6</span><br><span class="line-number">7</span><br><span class="line-number">8</span><br></div></div><p><strong>示例 2：</strong></p> <div class="language- line-numbers-mode"><pre class="language-text"><code>输入：nums = [0,1,1]
输出：[]
解释：唯一可能的三元组和不为 0 。
</code></pre> <div class="line-numbers-wrapper"><span class="line-number">1</span><br><span class="line-number">2</span><br><span class="line-number">3</span><br></div></div><p><strong>示例 3：</strong></p> <div class="language- line-numbers-mode"><pre class="language-text"><code>输入：nums = [0,0,0]
输出：[[0,0,0]]
解释：唯一可能的三元组和为 0 。
</code></pre> <div class="line-numbers-wrapper"><span class="line-number">1</span><br><span class="line-number">2</span><br><span class="line-number">3</span><br></div></div><p><strong>提示：</strong></p> <ul><li><code>3 &lt;= nums.length &lt;= 3000</code></li> <li><code>-105 &lt;= nums[i] &lt;= 105</code></li></ul> <h5 id="解题思路-2"><a href="#解题思路-2" class="header-anchor">#</a> 解题思路</h5> <div class="language-shell line-numbers-mode"><pre class="language-shell"><code><span class="token number">1</span>. 三数之和，转换为两个两数之和
</code></pre> <div class="line-numbers-wrapper"><span class="line-number">1</span><br></div></div><h5 id="完整代码-2"><a href="#完整代码-2" class="header-anchor">#</a> 完整代码</h5> <div class="language-js line-numbers-mode"><pre class="language-js"><code><span class="token comment">/**
 * @param {number[]} nums
 * @return {number[][]}
 */</span>
<span class="token keyword">var</span> <span class="token function-variable function">threeSum</span> <span class="token operator">=</span> <span class="token keyword">function</span><span class="token punctuation">(</span><span class="token parameter">nums</span><span class="token punctuation">)</span> <span class="token punctuation">{</span>

    <span class="token comment">// 解题思路</span>

    <span class="token comment">// 转换为两个两个之和</span>

    <span class="token comment">// nums[i]+nums[j]+nums[k]===0</span>
    <span class="token comment">// nums[j]+nums[k]===-nums[i]</span>

    <span class="token comment">// 先对nums进行排序，因为后面的两数之和算法需要有序</span>
    nums<span class="token punctuation">.</span><span class="token function">sort</span><span class="token punctuation">(</span><span class="token punctuation">(</span><span class="token parameter">a<span class="token punctuation">,</span>b</span><span class="token punctuation">)</span><span class="token operator">=&gt;</span>a<span class="token operator">-</span>b<span class="token punctuation">)</span>

    <span class="token comment">// console.log(nums)</span>

    <span class="token keyword">let</span> res<span class="token operator">=</span><span class="token punctuation">[</span><span class="token punctuation">]</span>

    <span class="token keyword">for</span><span class="token punctuation">(</span><span class="token keyword">let</span> i<span class="token operator">=</span><span class="token number">0</span><span class="token punctuation">;</span>i<span class="token operator">&lt;</span>nums<span class="token punctuation">.</span>length<span class="token punctuation">;</span>i<span class="token operator">++</span><span class="token punctuation">)</span><span class="token punctuation">{</span>
        <span class="token comment">// start为开始下标，因为i&lt;j&lt;k</span>
        <span class="token comment">// 有序数组的去重 ==&gt; nums[i] 相等时只需要计算一次</span>
        <span class="token keyword">if</span><span class="token punctuation">(</span>i<span class="token operator">&gt;</span><span class="token number">0</span><span class="token operator">&amp;&amp;</span>nums<span class="token punctuation">[</span>i<span class="token punctuation">]</span><span class="token operator">===</span>nums<span class="token punctuation">[</span>i<span class="token operator">-</span><span class="token number">1</span><span class="token punctuation">]</span><span class="token punctuation">)</span> <span class="token keyword">continue</span>
        <span class="token keyword">let</span> result<span class="token operator">=</span><span class="token function">twoSum</span><span class="token punctuation">(</span>nums<span class="token punctuation">,</span>i<span class="token operator">+</span><span class="token number">1</span><span class="token punctuation">,</span><span class="token operator">-</span>nums<span class="token punctuation">[</span>i<span class="token punctuation">]</span><span class="token punctuation">)</span>

        result<span class="token punctuation">.</span><span class="token function">forEach</span><span class="token punctuation">(</span><span class="token parameter">item</span><span class="token operator">=&gt;</span><span class="token punctuation">{</span>
            res<span class="token punctuation">.</span><span class="token function">push</span><span class="token punctuation">(</span>item<span class="token punctuation">)</span>
        <span class="token punctuation">}</span><span class="token punctuation">)</span>
    <span class="token punctuation">}</span>

    <span class="token keyword">return</span> res


<span class="token punctuation">}</span><span class="token punctuation">;</span>

<span class="token keyword">var</span> <span class="token function-variable function">twoSum</span> <span class="token operator">=</span> <span class="token keyword">function</span><span class="token punctuation">(</span><span class="token parameter">numbers<span class="token punctuation">,</span> start<span class="token punctuation">,</span>target</span><span class="token punctuation">)</span> <span class="token punctuation">{</span>

    <span class="token comment">// 使用双指针扫描</span>

    <span class="token comment">// 1. 滑动窗口 i,j方向相同</span>

    <span class="token comment">// 2 .i，j相向</span>

    <span class="token comment">// 这里 i，j相向</span>

    <span class="token comment">// 因为数组升序，i一开始最小，j一开始最大</span>
    <span class="token comment">// i+j&gt;target i不动，j左移 ==&gt; target 变小  ，如果满足，直接返回，如果&lt;target 说明i要右移 target变大</span>

    <span class="token comment">// 夹逼最后找到解</span>

    <span class="token keyword">let</span> j<span class="token operator">=</span>numbers<span class="token punctuation">.</span>length<span class="token operator">-</span><span class="token number">1</span>
    <span class="token keyword">let</span> res<span class="token operator">=</span><span class="token punctuation">[</span><span class="token punctuation">]</span>
    <span class="token keyword">for</span><span class="token punctuation">(</span><span class="token keyword">let</span> i<span class="token operator">=</span>start<span class="token punctuation">;</span>i<span class="token operator">&lt;</span>numbers<span class="token punctuation">.</span>length<span class="token punctuation">;</span>i<span class="token operator">++</span><span class="token punctuation">)</span><span class="token punctuation">{</span>

        <span class="token comment">// 这里也需要去重，即nums[j] 相同时，没有必要重复计算</span>
        <span class="token keyword">if</span><span class="token punctuation">(</span>i<span class="token operator">&gt;</span>start<span class="token operator">&amp;&amp;</span>numbers<span class="token punctuation">[</span>i<span class="token punctuation">]</span><span class="token operator">===</span>numbers<span class="token punctuation">[</span>i<span class="token operator">-</span><span class="token number">1</span><span class="token punctuation">]</span><span class="token punctuation">)</span> <span class="token keyword">continue</span>
        <span class="token keyword">while</span><span class="token punctuation">(</span>i<span class="token operator">&lt;</span>j<span class="token operator">&amp;&amp;</span>numbers<span class="token punctuation">[</span>i<span class="token punctuation">]</span><span class="token operator">+</span>numbers<span class="token punctuation">[</span>j<span class="token punctuation">]</span><span class="token operator">&gt;</span>target<span class="token punctuation">)</span><span class="token punctuation">{</span>
            j<span class="token operator">--</span>
        <span class="token punctuation">}</span>

        <span class="token keyword">if</span><span class="token punctuation">(</span>i<span class="token operator">&lt;</span>j<span class="token operator">&amp;&amp;</span>numbers<span class="token punctuation">[</span>i<span class="token punctuation">]</span><span class="token operator">+</span>numbers<span class="token punctuation">[</span>j<span class="token punctuation">]</span><span class="token operator">===</span>target<span class="token punctuation">)</span><span class="token punctuation">{</span>
            <span class="token comment">// 这里返回的为数组的数组，存储nums[i] 固定时，j，k的组合</span>
            <span class="token comment">// return [i+1,j+1]</span>
            res<span class="token punctuation">.</span><span class="token function">push</span><span class="token punctuation">(</span><span class="token punctuation">[</span><span class="token operator">-</span>target<span class="token punctuation">,</span>numbers<span class="token punctuation">[</span>i<span class="token punctuation">]</span><span class="token punctuation">,</span>numbers<span class="token punctuation">[</span>j<span class="token punctuation">]</span><span class="token punctuation">]</span><span class="token punctuation">)</span>
        <span class="token punctuation">}</span>
    <span class="token punctuation">}</span>

    <span class="token keyword">return</span> res

<span class="token punctuation">}</span><span class="token punctuation">;</span>
</code></pre> <div class="line-numbers-wrapper"><span class="line-number">1</span><br><span class="line-number">2</span><br><span class="line-number">3</span><br><span class="line-number">4</span><br><span class="line-number">5</span><br><span class="line-number">6</span><br><span class="line-number">7</span><br><span class="line-number">8</span><br><span class="line-number">9</span><br><span class="line-number">10</span><br><span class="line-number">11</span><br><span class="line-number">12</span><br><span class="line-number">13</span><br><span class="line-number">14</span><br><span class="line-number">15</span><br><span class="line-number">16</span><br><span class="line-number">17</span><br><span class="line-number">18</span><br><span class="line-number">19</span><br><span class="line-number">20</span><br><span class="line-number">21</span><br><span class="line-number">22</span><br><span class="line-number">23</span><br><span class="line-number">24</span><br><span class="line-number">25</span><br><span class="line-number">26</span><br><span class="line-number">27</span><br><span class="line-number">28</span><br><span class="line-number">29</span><br><span class="line-number">30</span><br><span class="line-number">31</span><br><span class="line-number">32</span><br><span class="line-number">33</span><br><span class="line-number">34</span><br><span class="line-number">35</span><br><span class="line-number">36</span><br><span class="line-number">37</span><br><span class="line-number">38</span><br><span class="line-number">39</span><br><span class="line-number">40</span><br><span class="line-number">41</span><br><span class="line-number">42</span><br><span class="line-number">43</span><br><span class="line-number">44</span><br><span class="line-number">45</span><br><span class="line-number">46</span><br><span class="line-number">47</span><br><span class="line-number">48</span><br><span class="line-number">49</span><br><span class="line-number">50</span><br><span class="line-number">51</span><br><span class="line-number">52</span><br><span class="line-number">53</span><br><span class="line-number">54</span><br><span class="line-number">55</span><br><span class="line-number">56</span><br><span class="line-number">57</span><br><span class="line-number">58</span><br><span class="line-number">59</span><br><span class="line-number">60</span><br><span class="line-number">61</span><br><span class="line-number">62</span><br><span class="line-number">63</span><br><span class="line-number">64</span><br><span class="line-number">65</span><br><span class="line-number">66</span><br><span class="line-number">67</span><br><span class="line-number">68</span><br><span class="line-number">69</span><br><span class="line-number">70</span><br><span class="line-number">71</span><br></div></div><ul><li>注意点
<ul><li>需要对数组进行排序</li> <li>需要去除重复项</li></ul></li></ul> <h4 id="_11-盛最多水的容器"><a href="#_11-盛最多水的容器" class="header-anchor">#</a> <a href="https://leetcode.cn/problems/container-with-most-water/description/" target="_blank" rel="noopener noreferrer">11. 盛最多水的容器<span><svg xmlns="http://www.w3.org/2000/svg" aria-hidden="true" focusable="false" x="0px" y="0px" viewBox="0 0 100 100" width="15" height="15" class="icon outbound"><path fill="currentColor" d="M18.8,85.1h56l0,0c2.2,0,4-1.8,4-4v-32h-8v28h-48v-48h28v-8h-32l0,0c-2.2,0-4,1.8-4,4v56C14.8,83.3,16.6,85.1,18.8,85.1z"></path> <polygon fill="currentColor" points="45.7,48.7 51.3,54.3 77.2,28.5 77.2,37.2 85.2,37.2 85.2,14.9 62.8,14.9 62.8,22.9 71.5,22.9"></polygon></svg> <span class="sr-only">(opens new window)</span></span></a></h4> <p>给定一个长度为 <code>n</code> 的整数数组 <code>height</code> 。有 <code>n</code> 条垂线，第 <code>i</code> 条线的两个端点是 <code>(i, 0)</code> 和 <code>(i, height[i])</code> 。</p> <p>找出其中的两条线，使得它们与 <code>x</code> 轴共同构成的容器可以容纳最多的水。</p> <p>返回容器可以储存的最大水量。</p> <p>**说明：**你不能倾斜容器。</p> <p><strong>示例 1：</strong></p> <p><img src="https://aliyun-lc-upload.oss-cn-hangzhou.aliyuncs.com/aliyun-lc-upload/uploads/2018/07/25/question_11.jpg" alt="img"></p> <div class="language- line-numbers-mode"><pre class="language-text"><code>输入：[1,8,6,2,5,4,8,3,7]
输出：49 
解释：图中垂直线代表输入数组 [1,8,6,2,5,4,8,3,7]。在此情况下，容器能够容纳水（表示为蓝色部分）的最大值为 49。
</code></pre> <div class="line-numbers-wrapper"><span class="line-number">1</span><br><span class="line-number">2</span><br><span class="line-number">3</span><br></div></div><p><strong>示例 2：</strong></p> <div class="language- line-numbers-mode"><pre class="language-text"><code>输入：height = [1,1]
输出：1
</code></pre> <div class="line-numbers-wrapper"><span class="line-number">1</span><br><span class="line-number">2</span><br></div></div><p><strong>提示：</strong></p> <ul><li><code>n == height.length</code></li> <li><code>2 &lt;= n &lt;= 105</code></li> <li><code>0 &lt;= height[i] &lt;= 104</code></li></ul> <h5 id="完整代码-3"><a href="#完整代码-3" class="header-anchor">#</a> 完整代码</h5> <div class="language-js line-numbers-mode"><pre class="language-js"><code><span class="token comment">/**
 * @param {number[]} height
 * @return {number}
 */</span>
<span class="token keyword">var</span> <span class="token function-variable function">maxArea</span> <span class="token operator">=</span> <span class="token keyword">function</span><span class="token punctuation">(</span><span class="token parameter">height</span><span class="token punctuation">)</span> <span class="token punctuation">{</span>

    <span class="token comment">// 解题思路</span>

    <span class="token comment">// 双指针夹逼</span>

    <span class="token comment">// 木桶水的容量由短的那块木板决定</span>

    <span class="token comment">// 刚开始 res=(j-i)*h[短]</span>

    <span class="token comment">// 如果长的木板舍弃掉，往里夹逼，容量一定减小（j-i变小，而短木板没动） </span>
    <span class="token comment">// 如果短木板舍弃，往里夹逼，（j-i变小） ，容量可能增大（短的木板下一位可能比之前的长，上限变高）</span>

    <span class="token comment">// 所以我们每次舍弃短的木板，求哪个最大</span>

    
    <span class="token keyword">let</span> i<span class="token operator">=</span><span class="token number">0</span>
    <span class="token keyword">let</span> j<span class="token operator">=</span>height<span class="token punctuation">.</span>length<span class="token operator">-</span><span class="token number">1</span>

    <span class="token keyword">let</span> max<span class="token operator">=</span><span class="token number">0</span>
    <span class="token keyword">while</span><span class="token punctuation">(</span>i<span class="token operator">&lt;</span>j<span class="token punctuation">)</span><span class="token punctuation">{</span>

        <span class="token comment">// 短的木板内移</span>
        <span class="token keyword">if</span><span class="token punctuation">(</span>height<span class="token punctuation">[</span>i<span class="token punctuation">]</span><span class="token operator">&lt;</span>height<span class="token punctuation">[</span>j<span class="token punctuation">]</span><span class="token punctuation">)</span><span class="token punctuation">{</span>
            max<span class="token operator">=</span>Math<span class="token punctuation">.</span><span class="token function">max</span><span class="token punctuation">(</span>max<span class="token punctuation">,</span>height<span class="token punctuation">[</span>i<span class="token punctuation">]</span><span class="token operator">*</span><span class="token punctuation">(</span>j<span class="token operator">-</span>i<span class="token punctuation">)</span><span class="token punctuation">)</span>
            i<span class="token operator">++</span>
        <span class="token punctuation">}</span><span class="token keyword">else</span><span class="token punctuation">{</span>
            max<span class="token operator">=</span>Math<span class="token punctuation">.</span><span class="token function">max</span><span class="token punctuation">(</span>max<span class="token punctuation">,</span>height<span class="token punctuation">[</span>j<span class="token punctuation">]</span><span class="token operator">*</span><span class="token punctuation">(</span>j<span class="token operator">-</span>i<span class="token punctuation">)</span><span class="token punctuation">)</span>
            j<span class="token operator">--</span>
        <span class="token punctuation">}</span>
    <span class="token punctuation">}</span>

    <span class="token keyword">return</span> max

<span class="token punctuation">}</span><span class="token punctuation">;</span>
</code></pre> <div class="line-numbers-wrapper"><span class="line-number">1</span><br><span class="line-number">2</span><br><span class="line-number">3</span><br><span class="line-number">4</span><br><span class="line-number">5</span><br><span class="line-number">6</span><br><span class="line-number">7</span><br><span class="line-number">8</span><br><span class="line-number">9</span><br><span class="line-number">10</span><br><span class="line-number">11</span><br><span class="line-number">12</span><br><span class="line-number">13</span><br><span class="line-number">14</span><br><span class="line-number">15</span><br><span class="line-number">16</span><br><span class="line-number">17</span><br><span class="line-number">18</span><br><span class="line-number">19</span><br><span class="line-number">20</span><br><span class="line-number">21</span><br><span class="line-number">22</span><br><span class="line-number">23</span><br><span class="line-number">24</span><br><span class="line-number">25</span><br><span class="line-number">26</span><br><span class="line-number">27</span><br><span class="line-number">28</span><br><span class="line-number">29</span><br><span class="line-number">30</span><br><span class="line-number">31</span><br><span class="line-number">32</span><br><span class="line-number">33</span><br><span class="line-number">34</span><br><span class="line-number">35</span><br><span class="line-number">36</span><br><span class="line-number">37</span><br><span class="line-number">38</span><br><span class="line-number">39</span><br></div></div><h4 id="_84-柱状图中最大的矩形-单调栈模板"><a href="#_84-柱状图中最大的矩形-单调栈模板" class="header-anchor">#</a> <a href="https://leetcode.cn/problems/largest-rectangle-in-histogram/description/" target="_blank" rel="noopener noreferrer">84. 柱状图中最大的矩形(单调栈模板)<span><svg xmlns="http://www.w3.org/2000/svg" aria-hidden="true" focusable="false" x="0px" y="0px" viewBox="0 0 100 100" width="15" height="15" class="icon outbound"><path fill="currentColor" d="M18.8,85.1h56l0,0c2.2,0,4-1.8,4-4v-32h-8v28h-48v-48h28v-8h-32l0,0c-2.2,0-4,1.8-4,4v56C14.8,83.3,16.6,85.1,18.8,85.1z"></path> <polygon fill="currentColor" points="45.7,48.7 51.3,54.3 77.2,28.5 77.2,37.2 85.2,37.2 85.2,14.9 62.8,14.9 62.8,22.9 71.5,22.9"></polygon></svg> <span class="sr-only">(opens new window)</span></span></a></h4> <p>给定 <em>n</em> 个非负整数，用来表示柱状图中各个柱子的高度。每个柱子彼此相邻，且宽度为 1 。</p> <p>求在该柱状图中，能够勾勒出来的矩形的最大面积。</p> <p><strong>示例 1:</strong></p> <p><img src="https://assets.leetcode.com/uploads/2021/01/04/histogram.jpg" alt="img"></p> <div class="language- line-numbers-mode"><pre class="language-text"><code>输入：heights = [2,1,5,6,2,3]
输出：10
解释：最大的矩形为图中红色区域，面积为 10
</code></pre> <div class="line-numbers-wrapper"><span class="line-number">1</span><br><span class="line-number">2</span><br><span class="line-number">3</span><br></div></div><p><strong>示例 2：</strong></p> <p><img src="https://assets.leetcode.com/uploads/2021/01/04/histogram-1.jpg" alt="img"></p> <div class="language- line-numbers-mode"><pre class="language-text"><code>输入： heights = [2,4]
输出： 4
</code></pre> <div class="line-numbers-wrapper"><span class="line-number">1</span><br><span class="line-number">2</span><br></div></div><p><strong>提示：</strong></p> <ul><li><code>1 &lt;= heights.length &lt;=105</code></li> <li><code>0 &lt;= heights[i] &lt;= 104</code></li></ul> <h5 id="解题思路-3"><a href="#解题思路-3" class="header-anchor">#</a> 解题思路</h5> <div class="language-shell line-numbers-mode"><pre class="language-shell"><code><span class="token number">1</span>.可以使用暴力枚举，但是O<span class="token punctuation">(</span>n*b<span class="token punctuation">)</span>
	从当前矩形往左右扩散
	可以相连则计算一下最值
	
	循环全部矩形，重复上述操作
	
<span class="token number">2</span>. 单调栈解法
	<span class="token keyword">for</span> 矩形
	保留递增的矩形（入栈）
	遇到递减的（计算递增矩形序列中的最值，pop递增矩形）
	插入该递减的值
	
	重新构成了递增的矩形序列，重复上述操作
</code></pre> <div class="line-numbers-wrapper"><span class="line-number">1</span><br><span class="line-number">2</span><br><span class="line-number">3</span><br><span class="line-number">4</span><br><span class="line-number">5</span><br><span class="line-number">6</span><br><span class="line-number">7</span><br><span class="line-number">8</span><br><span class="line-number">9</span><br><span class="line-number">10</span><br><span class="line-number">11</span><br><span class="line-number">12</span><br><span class="line-number">13</span><br></div></div><ul><li>计算递增矩形序列最值</li></ul> <p><img src="https://img2023.cnblogs.com/blog/3089561/202302/3089561-20230203130631629-1925052882.jpg" alt=""></p> <h5 id="完整代码-4"><a href="#完整代码-4" class="header-anchor">#</a> 完整代码</h5> <div class="language-js line-numbers-mode"><pre class="language-js"><code><span class="token comment">/**
 * @param {number[]} heights
 * @return {number}
 */</span>
<span class="token keyword">var</span> <span class="token function-variable function">largestRectangleArea</span> <span class="token operator">=</span> <span class="token keyword">function</span><span class="token punctuation">(</span><span class="token parameter">heights</span><span class="token punctuation">)</span> <span class="token punctuation">{</span>

    <span class="token comment">// 单调栈解决</span>

    <span class="token comment">// 如果后面的数递增，入栈，</span>

    <span class="token comment">// 遇到递减的，我们开始计算之前单调栈的最值，pop值，直到小于新的值，新值入栈</span>

    <span class="token comment">// 再次变成了单调栈，重复上述操作</span>

    <span class="token keyword">let</span> stack<span class="token operator">=</span><span class="token punctuation">[</span><span class="token punctuation">]</span>

    <span class="token comment">// 写一个临界条件</span>
    <span class="token comment">// 最后一个值为0</span>
    <span class="token comment">// 用于终止递增的单调栈</span>

    heights<span class="token punctuation">.</span><span class="token function">push</span><span class="token punctuation">(</span><span class="token number">0</span><span class="token punctuation">)</span>

    <span class="token keyword">let</span> res<span class="token operator">=</span><span class="token number">0</span>
    <span class="token keyword">for</span><span class="token punctuation">(</span><span class="token keyword">let</span> i<span class="token operator">=</span><span class="token number">0</span><span class="token punctuation">;</span>i<span class="token operator">&lt;</span>heights<span class="token punctuation">.</span>length<span class="token punctuation">;</span>i<span class="token operator">++</span><span class="token punctuation">)</span><span class="token punctuation">{</span>

        <span class="token keyword">let</span> account_width<span class="token operator">=</span><span class="token number">0</span>
        <span class="token comment">// console.log(heights[i])</span>
        <span class="token comment">// 当不递增时</span>
        <span class="token keyword">while</span><span class="token punctuation">(</span>stack<span class="token punctuation">.</span>length<span class="token operator">&amp;&amp;</span><span class="token punctuation">(</span>stack<span class="token punctuation">[</span>stack<span class="token punctuation">.</span>length<span class="token operator">-</span><span class="token number">1</span><span class="token punctuation">]</span><span class="token punctuation">.</span>height<span class="token operator">&gt;=</span>heights<span class="token punctuation">[</span>i<span class="token punctuation">]</span><span class="token punctuation">)</span><span class="token punctuation">)</span><span class="token punctuation">{</span>
            <span class="token comment">// 从最高点的开始往左扩散</span>
            <span class="token comment">// width上升</span>
            account_width<span class="token operator">+=</span>stack<span class="token punctuation">[</span>stack<span class="token punctuation">.</span>length<span class="token operator">-</span><span class="token number">1</span><span class="token punctuation">]</span><span class="token punctuation">.</span>width
            <span class="token comment">// console.log(stack)</span>
            res<span class="token operator">=</span>Math<span class="token punctuation">.</span><span class="token function">max</span><span class="token punctuation">(</span>res<span class="token punctuation">,</span>account_width<span class="token operator">*</span>stack<span class="token punctuation">[</span>stack<span class="token punctuation">.</span>length<span class="token operator">-</span><span class="token number">1</span><span class="token punctuation">]</span><span class="token punctuation">.</span>height<span class="token punctuation">)</span>
            stack<span class="token punctuation">.</span><span class="token function">pop</span><span class="token punctuation">(</span><span class="token punctuation">)</span>
        <span class="token punctuation">}</span>

        <span class="token comment">// 递增，入栈</span>
        <span class="token comment">// 这里account_width 要加上去，因为原来在栈中的被pop出去了</span>
        stack<span class="token punctuation">.</span><span class="token function">push</span><span class="token punctuation">(</span><span class="token punctuation">{</span><span class="token literal-property property">width</span><span class="token operator">:</span>account_width<span class="token operator">+</span><span class="token number">1</span><span class="token punctuation">,</span><span class="token literal-property property">height</span><span class="token operator">:</span>heights<span class="token punctuation">[</span>i<span class="token punctuation">]</span><span class="token punctuation">}</span><span class="token punctuation">)</span>
    <span class="token punctuation">}</span>

    <span class="token keyword">return</span> res
<span class="token punctuation">}</span><span class="token punctuation">;</span>
</code></pre> <div class="line-numbers-wrapper"><span class="line-number">1</span><br><span class="line-number">2</span><br><span class="line-number">3</span><br><span class="line-number">4</span><br><span class="line-number">5</span><br><span class="line-number">6</span><br><span class="line-number">7</span><br><span class="line-number">8</span><br><span class="line-number">9</span><br><span class="line-number">10</span><br><span class="line-number">11</span><br><span class="line-number">12</span><br><span class="line-number">13</span><br><span class="line-number">14</span><br><span class="line-number">15</span><br><span class="line-number">16</span><br><span class="line-number">17</span><br><span class="line-number">18</span><br><span class="line-number">19</span><br><span class="line-number">20</span><br><span class="line-number">21</span><br><span class="line-number">22</span><br><span class="line-number">23</span><br><span class="line-number">24</span><br><span class="line-number">25</span><br><span class="line-number">26</span><br><span class="line-number">27</span><br><span class="line-number">28</span><br><span class="line-number">29</span><br><span class="line-number">30</span><br><span class="line-number">31</span><br><span class="line-number">32</span><br><span class="line-number">33</span><br><span class="line-number">34</span><br><span class="line-number">35</span><br><span class="line-number">36</span><br><span class="line-number">37</span><br><span class="line-number">38</span><br><span class="line-number">39</span><br><span class="line-number">40</span><br><span class="line-number">41</span><br><span class="line-number">42</span><br><span class="line-number">43</span><br><span class="line-number">44</span><br></div></div><h4 id="_239-滑动窗口最大值-单调队列模板题"><a href="#_239-滑动窗口最大值-单调队列模板题" class="header-anchor">#</a> <a href="https://leetcode.cn/problems/sliding-window-maximum/description/" target="_blank" rel="noopener noreferrer">239. 滑动窗口最大值(单调队列模板题)<span><svg xmlns="http://www.w3.org/2000/svg" aria-hidden="true" focusable="false" x="0px" y="0px" viewBox="0 0 100 100" width="15" height="15" class="icon outbound"><path fill="currentColor" d="M18.8,85.1h56l0,0c2.2,0,4-1.8,4-4v-32h-8v28h-48v-48h28v-8h-32l0,0c-2.2,0-4,1.8-4,4v56C14.8,83.3,16.6,85.1,18.8,85.1z"></path> <polygon fill="currentColor" points="45.7,48.7 51.3,54.3 77.2,28.5 77.2,37.2 85.2,37.2 85.2,14.9 62.8,14.9 62.8,22.9 71.5,22.9"></polygon></svg> <span class="sr-only">(opens new window)</span></span></a></h4> <p>给你一个整数数组 <code>nums</code>，有一个大小为 <code>k</code> 的滑动窗口从数组的最左侧移动到数组的最右侧。你只可以看到在滑动窗口内的 <code>k</code> 个数字。滑动窗口每次只向右移动一位。</p> <p>返回 <em>滑动窗口中的最大值</em> 。</p> <p><strong>示例 1：</strong></p> <div class="language- line-numbers-mode"><pre class="language-text"><code>输入：nums = [1,3,-1,-3,5,3,6,7], k = 3
输出：[3,3,5,5,6,7]
解释：
滑动窗口的位置                最大值
---------------               -----
[1  3  -1] -3  5  3  6  7       3
 1 [3  -1  -3] 5  3  6  7       3
 1  3 [-1  -3  5] 3  6  7       5
 1  3  -1 [-3  5  3] 6  7       5
 1  3  -1  -3 [5  3  6] 7       6
 1  3  -1  -3  5 [3  6  7]      7
</code></pre> <div class="line-numbers-wrapper"><span class="line-number">1</span><br><span class="line-number">2</span><br><span class="line-number">3</span><br><span class="line-number">4</span><br><span class="line-number">5</span><br><span class="line-number">6</span><br><span class="line-number">7</span><br><span class="line-number">8</span><br><span class="line-number">9</span><br><span class="line-number">10</span><br><span class="line-number">11</span><br></div></div><p><strong>示例 2：</strong></p> <div class="language- line-numbers-mode"><pre class="language-text"><code>输入：nums = [1], k = 1
输出：[1]
</code></pre> <div class="line-numbers-wrapper"><span class="line-number">1</span><br><span class="line-number">2</span><br></div></div><p><strong>提示：</strong></p> <ul><li><code>1 &lt;= nums.length &lt;= 105</code></li> <li><code>-104 &lt;= nums[i] &lt;= 104</code></li> <li><code>1 &lt;= k &lt;= nums.length</code></li></ul> <h5 id="完整代码-5"><a href="#完整代码-5" class="header-anchor">#</a> 完整代码</h5> <div class="language-js line-numbers-mode"><pre class="language-js"><code><span class="token comment">/**
 * @param {number[]} nums
 * @param {number} k
 * @return {number[]}
 */</span>
<span class="token keyword">var</span> <span class="token function-variable function">maxSlidingWindow</span> <span class="token operator">=</span> <span class="token keyword">function</span><span class="token punctuation">(</span><span class="token parameter">nums<span class="token punctuation">,</span> k</span><span class="token punctuation">)</span> <span class="token punctuation">{</span>

    <span class="token comment">// 解题思路</span>

    <span class="token comment">// 维护一个单调递减的队列</span>

    <span class="token comment">// 因为后面插入的数如果大于前面的,说明前面的数是不是没用了，当前窗口的最大值为后插入的数</span>

    <span class="token comment">// 所以pop 前面的数，直到递减序列，这样可以保证队头为最大值</span>

    <span class="token comment">// 同时队头可能越界，不在窗口中，需要出队</span>

    <span class="token keyword">let</span> deQueue<span class="token operator">=</span><span class="token punctuation">[</span><span class="token punctuation">]</span>

    <span class="token keyword">let</span> res<span class="token operator">=</span><span class="token punctuation">[</span><span class="token punctuation">]</span>
    <span class="token keyword">for</span><span class="token punctuation">(</span><span class="token keyword">let</span> i<span class="token operator">=</span><span class="token number">0</span><span class="token punctuation">;</span>i<span class="token operator">&lt;</span>nums<span class="token punctuation">.</span>length<span class="token punctuation">;</span>i<span class="token operator">++</span><span class="token punctuation">)</span><span class="token punctuation">{</span>
        <span class="token comment">// 需要判断队头是否出界了</span>
        <span class="token comment">// dequeue存放的为下标，i-k为滑动串口的起点</span>
        <span class="token keyword">while</span><span class="token punctuation">(</span>deQueue<span class="token punctuation">.</span>length<span class="token operator">&amp;&amp;</span> deQueue<span class="token punctuation">[</span><span class="token number">0</span><span class="token punctuation">]</span><span class="token operator">&lt;=</span>i<span class="token operator">-</span>k<span class="token punctuation">)</span> deQueue<span class="token punctuation">.</span><span class="token function">shift</span><span class="token punctuation">(</span><span class="token punctuation">)</span>

        <span class="token comment">// 存放单调递减的序列</span>
        <span class="token keyword">while</span><span class="token punctuation">(</span>deQueue<span class="token punctuation">.</span>length<span class="token operator">&amp;&amp;</span> nums<span class="token punctuation">[</span>deQueue<span class="token punctuation">[</span>deQueue<span class="token punctuation">.</span>length<span class="token operator">-</span><span class="token number">1</span><span class="token punctuation">]</span><span class="token punctuation">]</span><span class="token operator">&lt;=</span>nums<span class="token punctuation">[</span>i<span class="token punctuation">]</span><span class="token punctuation">)</span> deQueue<span class="token punctuation">.</span><span class="token function">pop</span><span class="token punctuation">(</span><span class="token punctuation">)</span>

        deQueue<span class="token punctuation">.</span><span class="token function">push</span><span class="token punctuation">(</span>i<span class="token punctuation">)</span>

        <span class="token keyword">if</span><span class="token punctuation">(</span>i<span class="token operator">&gt;=</span>k<span class="token operator">-</span><span class="token number">1</span><span class="token punctuation">)</span><span class="token punctuation">{</span>
            res<span class="token punctuation">.</span><span class="token function">push</span><span class="token punctuation">(</span>nums<span class="token punctuation">[</span>deQueue<span class="token punctuation">[</span><span class="token number">0</span><span class="token punctuation">]</span><span class="token punctuation">]</span><span class="token punctuation">)</span>
        <span class="token punctuation">}</span>
        

    <span class="token punctuation">}</span>

    <span class="token keyword">return</span> res

    

<span class="token punctuation">}</span><span class="token punctuation">;</span>
</code></pre> <div class="line-numbers-wrapper"><span class="line-number">1</span><br><span class="line-number">2</span><br><span class="line-number">3</span><br><span class="line-number">4</span><br><span class="line-number">5</span><br><span class="line-number">6</span><br><span class="line-number">7</span><br><span class="line-number">8</span><br><span class="line-number">9</span><br><span class="line-number">10</span><br><span class="line-number">11</span><br><span class="line-number">12</span><br><span class="line-number">13</span><br><span class="line-number">14</span><br><span class="line-number">15</span><br><span class="line-number">16</span><br><span class="line-number">17</span><br><span class="line-number">18</span><br><span class="line-number">19</span><br><span class="line-number">20</span><br><span class="line-number">21</span><br><span class="line-number">22</span><br><span class="line-number">23</span><br><span class="line-number">24</span><br><span class="line-number">25</span><br><span class="line-number">26</span><br><span class="line-number">27</span><br><span class="line-number">28</span><br><span class="line-number">29</span><br><span class="line-number">30</span><br><span class="line-number">31</span><br><span class="line-number">32</span><br><span class="line-number">33</span><br><span class="line-number">34</span><br><span class="line-number">35</span><br><span class="line-number">36</span><br><span class="line-number">37</span><br><span class="line-number">38</span><br><span class="line-number">39</span><br><span class="line-number">40</span><br><span class="line-number">41</span><br><span class="line-number">42</span><br></div></div></div></div>  <div class="page-edit"><div class="edit-link"><a href="https://github.com/linxin1123/vuepress-blog/edit/master/docs/07.算法/41.算法训练营/05.前缀和-差分-双指针（下）-cnblog.md" target="_blank" rel="noopener noreferrer">编辑</a> <span><svg xmlns="http://www.w3.org/2000/svg" aria-hidden="true" focusable="false" x="0px" y="0px" viewBox="0 0 100 100" width="15" height="15" class="icon outbound"><path fill="currentColor" d="M18.8,85.1h56l0,0c2.2,0,4-1.8,4-4v-32h-8v28h-48v-48h28v-8h-32l0,0c-2.2,0-4,1.8-4,4v56C14.8,83.3,16.6,85.1,18.8,85.1z"></path> <polygon fill="currentColor" points="45.7,48.7 51.3,54.3 77.2,28.5 77.2,37.2 85.2,37.2 85.2,14.9 62.8,14.9 62.8,22.9 71.5,22.9"></polygon></svg> <span class="sr-only">(opens new window)</span></span></div> <!----> <div class="last-updated"><span class="prefix">上次更新:</span> <span class="time">2023/02/17, 11:29:32</span></div></div> <div class="page-nav-wapper"><div class="page-nav-centre-wrap"><a href="/pages/32c8c0/" class="page-nav-centre page-nav-centre-prev"><div class="tooltip">前缀和-差分-双指针（上）-cnblog</div></a> <a href="/pages/7f3ede/" class="page-nav-centre page-nav-centre-next"><div class="tooltip">算法训练营-排序-cnblog</div></a></div> <div class="page-nav"><p class="inner"><span class="prev">
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